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To evaluate the integral $I = \int_0^\infty \frac{\tan^{-1} (\ln x)}{1+x^2} dx$, I used a similar method to the one outlined in this post, where by using the substitution $u=\frac{1}{x}$ I found $I=-I$ and hence $I=0$.

It seems the only properties of $\tan^{-1} x$ used is that it is an odd function. However, according to WolframAlpha $\int_0^\infty \frac{f(\ln x)}{1+x^2} dx = 0$ for $f(x) = x, x^3, \tan^{-1} x, \tan x, \sin x, \sin^{-1} x, \sqrt[3] x$, and $\frac{1}{\sqrt[3] x}$ but not for $f(x) = \frac{1}{x}, \frac{1}{x^3}$ or $\sinh x$ as the integral does not converge, and WolframAlpha does not load an answer for $f(x) = \cot x$. Therefore, there must be some other condition limiting what odd functions $f$ can be chosen.

I suspect this condition may be related to whether the function is defined at $0$ or whether $\lim_{x \to \infty} f(x)$ exists, though there are counterexamples to both conditions individually. Any help is appreciated.

Chelsea
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2 Answers2

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If the integral converges and $f$ is odd, then it is $0$.

Proof: Rewrite the integral $$ \int_0^\infty \frac{f(\ln(x))}{1+x^2}dx =\int_0^\infty \frac{f(\ln(x))}{1/x+x}\frac{dx}x $$ Now substitute $u=\ln(x)$ and $du=\frac{dx}x$, so the integral becomes $$ \int_0^\infty \frac{f(\ln(x))}{1+x^2}dx= \int_{-\infty}^\infty \frac{f(u)}{e^{-u}+e^u}du $$ which is the integral of an odd function over a symmetric interval, hence it is $0$ if it converges. If it doesn't converge, as is the case with $f(x)=1/x$ or $\cot x$, then it just doesn't converge. But for odd $f$, it can't converge to anything nonzero.

As far as checking convergence is concerned, a sufficient condition is that $\int_0^\infty |f(x)|e^{-x}dx$ converges, because $\frac1{e^{-x}+e^x}\le e^{-x}$. On the other hand, the integral diverges if $\int_0^\epsilon f(x)dx$ is divergent for some $\epsilon>0$.

Dark Malthorp
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The problem can be generalized. Assume $f(x)$ is odd and $e^xg(e^x)$ is even. Equivalently $t^{-1}g(t^{-1})=tg(t).$ Then $$\int\limits_0^\infty f(\ln x)g(x)\,dx=0$$ Indeed, substitute $x=t^{-1}.$ Then $$\int\limits_0^\infty f(\ln x)g(x)\,dx= \int\limits_0^\infty f(-\ln t)t^{-2}g(t^{-1})\,dt = -\int\limits_0^\infty f(\ln t)g(t)\,dt $$ The function $g(x)=(1+x^2)^{-1}$ satisfies the assumptions.

Ryszard Szwarc
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